9.1/3575
use long division like this
_<u>0.</u>_________
3575|91.000000
we find how many of 3575 fit into the 91
it's too big so we go farther
how many 3575 go into 910
too big so we go farther
how many 3575 go into 9100
the answer is 2 so put that in the correct place and mulitply that and put that in the correct place. then we subtract
_<u>0.</u><u>02</u>_________
3575|91.000000
-<u>71.50</u>
19.50
bring down the next number
find how many go into 19.500
the answe ris 5
_<u>0.</u><u>02</u><u>5</u>_________
3575|91.000000
-<u>71.50</u>
19.500
-<u>17.875</u>
1.625
bring down he next zero ( I fast forward and skip steps for convinience)
__
_<u>0.</u><u>02</u><u>5</u><u>455</u>_________
3575|91.000000
-<u>71.50</u>
19.500
-<u>17.875</u>
1.6250
-<u>1.4300</u>
.19500
-<u>.17875
</u> 16250
-14300
and so on untill infinity so the answe ris 0.0254555555555555 (enless fives)
Step-by-step explanation:
step 1. x + 4.75 - 6.55 = 5.43
step 2. looks like b is the only answer.
Let 
be the number of days spent at Tahoe and San Francisco, respectively.
We don't know the values of 
and 
yet, but we know that the holiday lasted 9 days:
We also know that each day spent in Tahoe costed 350 and each day spent in San Francisco costed 475. So, the total cost of the holiday is the sum of the number of days muliplied by their cost:
If we put the two equations together, we have the system
Which yields
It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".
Anyway...
= = = = = = = = = = = = = = = = = =
The way you stated the problem, there is an infinity of possibilities for the other solution.
► For instance, the quadratic equation:
x² – (6 + 4i)x + (9 + 12i) = 0
has for discriminant:
Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16
which is indeed negative.
Its solutions will then be:
x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i
x₂ = [(6 + 4i) – 4i]/2 = 3
And the other solution here is 3.
► If you are not convinced, the quadratic equation:
x² – (6 + 5i)x + (5 + 15i) = 0
has for discriminant:
Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9
which is indeed negative.
Its solutions will then be:
x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i
x₂ = [(6 + 5i) – 3i]/2 = 3 + i
And the other solution here is 3+i.
► In fact, every quadratic equation of the form:
x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0
where α is any real, has for discriminant:
Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi)
= 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i
= 16α – (4 + α)²
= 16α – 16 – 8α – α²
= -16 + 8α – α²
= -(α – 4)²
WILL be negative.
Their solutions will then be:
x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i
x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi
And the other solution will then be is 3+αi.
Since α can take any real value, you'll obtain an infinity of solutions of the form 3+αi.
► So conclusively:
If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.
Regards, to lizard squad >:0
